Straight Lines And Pair Of Straight Lines – JEE Main MathematicsPractice Questions & PYQs

Minimizing the area of a triangle formed by a variable line passing through a fixed point relies on the AM-GM inequality. For a line passing through $(3,5)$ with intercepts $a$ and $b$, the equation $\frac{x}{a} + \frac{y}{b} = 1$ leads to $b = \frac{5a}{a-3}$ for $a > 3$. The area $\frac{1}{2}ab$ simplifies to $\frac{5}{2} \cdot \frac{a^2}{a-3}$, which can be expressed as $\frac{5}{2} \left( a+3 + \frac{9}{a-3} \right)$. By rearranging the term as $\frac{5}{2} \left( (a-3) + \frac{9}{a-3} + 6 \right)$, we apply the AM-GM inequality, where $(a-3) + \frac{9}{a-3} \ge 2\sqrt{(a-3) \cdot \frac{9}{a-3}} = 6$. The entire expression therefore reaches a minimum of $\frac{5}{2}(6 + 6)$, resulting in an area of 30.

Measuring the distance of a point from a line along a specific direction relies on the parametric form of a line equation. Moving from the point $(2, 3)$ at an angle $\theta$ for a distance $r$ means any point on this path is defined by the coordinates $(2+r\cos\theta, 3+r\sin\theta)$. Since the direction is parallel to the line $\sqrt{3}x - y + 1 = 0$, which has a slope of $\sqrt{3}$, we determine the angle $\theta$ by $\tan\theta = \sqrt{3}$. This yields $\theta = 60^\circ$, and consequently $\cos\theta = \frac{1}{2}$ and $\sin\theta = \frac{\sqrt{3}}{2}$.

Substituting these parameters into the coordinates gives $x = 2 + \frac{r}{2}$ and $y = 3 + \frac{\sqrt{3}r}{2}$. These coordinates must satisfy the line equation $2x - 3y + 28 = 0$ at the point of intersection. Substituting these expressions into the line equation leads to $2(2 + \frac{r}{2}) - 3(3 + \frac{\sqrt{3}r}{2}) + 28 = 0$. Expanding the terms results in $4 + r - 9 - \frac{3\sqrt{3}r}{2} + 28 = 0$, which simplifies to $23 + r - \frac{3\sqrt{3}r}{2} = 0$. Isolating $r$ through algebraic manipulation provides $r( \frac{3\sqrt{3}-2}{2} ) = 23$, which simplifies to $r = \frac{46}{3\sqrt{3}-2}$. Multiplying the numerator and denominator by the conjugate $3\sqrt{3}+2$ results in the distance of $4 + 6\sqrt{3}$.

The tangent of the angle between the lines $L_1$ and $L_2$ is derived by first identifying the x and y intercepts of the line $4x + 5y = 20$ in the first quadrant as $(5, 0)$ and $(0, 4)$. Dividing this segment into three equal parts using the section formula yields the points $A = \left( \frac{5}{3}, \frac{8}{3} \right)$ and $B = \left( \frac{10}{3}, \frac{4}{3} \right)$, which provide the slopes $m_1 = \frac{8/3}{5/3} = \frac{8}{5}$ and $m_2 = \frac{4/3}{10/3} = \frac{2}{5}$ for the two lines passing through the origin. Applying the formula for the angle $\theta$ between two lines, $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$, we substitute these slopes to find $\tan \theta = \left| \frac{8/5 - 2/5}{1 + (8/5)(2/5)} \right| = \frac{6/5}{1 + 16/25} = \frac{6/5}{41/25}$, which simplifies to $\frac{30}{41}$.

The centroid, circumcentre, and orthocentre of a triangle are collinear, with the centroid $A$ positioned such that it divides the segment connecting the orthocentre $C$ and the circumcentre $B$ in a $1:2$ ratio. Calculating this position using the coordinates $B(3,4)$ and $C(-6,-8)$ yields the centroid $A(0,0)$, which allows us to simplify the coordinates of $P(2a+3, 7b+5)$ to $(3,5)$.

With the point $P$ established, finding its distance from the line $2x+3y-4=0$ along the direction of $x-2y-1=0$ involves setting up a parametric line equation $x = 3 + r\cos\theta$ and $y = 5 + r\sin\theta$, where $\tan\theta = 1/2$ implies $\cos\theta = \frac{2}{\sqrt{5}}$ and $\sin\theta = \frac{1}{\sqrt{5}}$. Substituting these coordinates into the line equation $2x+3y-4=0$ leads to the expression $2(3 + \frac{2r}{\sqrt{5}}) + 3(5 + \frac{r}{\sqrt{5}}) - 4 = 0$. Solving for $r$, we find $17 + \frac{7r}{\sqrt{5}} = 0$, which results in a magnitude of $r = \frac{17\sqrt{5}}{7}$.

When a light ray reflects off a mirror, the path of the reflected ray, if traced back behind the surface, originates from the mirror image of the point through which the reflected light passes. Determining the reflection of the point $(7,2)$ across the line $2x+y-6=0$ requires locating the point $(x,y)$ such that the line connecting it to $(7,2)$ is perpendicular to the mirror and bisected by it. Applying the reflection formula $\frac{x-7}{2} = \frac{y-2}{1} = -2 \frac{2(7)+2-6}{2^2+1^2}$ provides the image coordinates; the expression simplifies to $\frac{x-7}{2} = \frac{y-2}{1} = -4$, resulting in the point $(-1,-2)$. Since the incident ray connects the initial point $(3,10)$ and this image point $(-1,-2)$, the slope is $\frac{10 - (-2)}{3 - (-1)} = 3$. The equation of this line in point-slope form is $y - 10 = 3(x - 3)$, which rearranges to $3x - y + 1 = 0$. Comparing this with $ax+by+1=0$ confirms $a=3$ and $b=-1$, which yields $a^2+b^2+3ab = 3^2+(-1)^2+3(3)(-1) = 9+1-9$, resulting in 1.

The core principle involves leveraging the parametric representation of points on the known sides $AB$ and $AC$ in conjunction with the section formula to isolate the specific coordinates of vertices $B$ and $C$. By expressing $B$ as $(x_1, 14-4x_1)$ and $C$ as $(x_2, \frac{3x_2-5}{2})$, the internal ratio $2:1$ for the point $(2, -\frac{4}{3})$ yields the system of equations $2x_2+x_1=6$ and $3x_2-4x_1=-13$. Solving this system identifies the vertices as $B(4, -2)$ and $C(1, -1)$, which defines a line with a slope of $-\frac{1}{3}$. Using the point-slope formula with these coordinates leads directly to the equation $x+3y+2=0$.

Three lines fail to form a triangle if they are concurrent-meeting at a single point-or if at least two of the lines are parallel to each other. Because the first two given lines, $2x - y + 3 = 0$ and $6x + 3y + 1 = 0$, possess slopes of $2$ and $-2$ respectively, they are not parallel to each other, meaning the third line, $\alpha x + 2y - 2 = 0$, must either be parallel to one of them or pass through their common intersection point. Setting the slope of the third line, $-\alpha/2$, equal to the slope of the first line gives $-\alpha/2 = 2$, or $\alpha = -4$, while setting it equal to the slope of the second line gives $-\alpha/2 = -2$, or $\alpha = 4$.

For the concurrency condition, the determinant of the coefficients of the three lines must vanish:

Expanding this determinant results in $2(3 \times -2 - 1 \times 2) - (-1)(6 \times -2 - 1 \times \alpha) + 3(6 \times 2 - 3 \times \alpha) = 0$, which simplifies to $2(-8) + 1(-12 - \alpha) + 3(12 - 3\alpha) = 0$. Further reduction yields $-16 - 12 - \alpha + 36 - 9\alpha = 0$, or $8 - 10\alpha = 0$, leading to $\alpha = 4/5$. Summing the squares of these three distinct values of $\alpha$ provides the value $p = (-4)^2 + 4^2 + (4/5)^2 = 16 + 16 + 16/25 = 32 + 0.64 = 32.64$. The greatest integer less than or equal to $p$ is 32.

The Angle Bisector Theorem establishes that the bisector of an interior angle of a triangle divides the opposite side into segments proportional to the lengths of the adjacent sides, meaning the ratio $AB/BC$ is equal to $AD/DC$. Identifying the intersection of the angle bisector $y = x$ and the side $AC$ given by $2x - y = 2$ provides the coordinates for point $D$, which is $(2, 2)$. Using the ratio $1:2$ and the section formula with point $A(4, 6)$ and point $D(2, 2)$, we determine the coordinates of vertex $C$ to be $(-2, -6)$. Since vertex $B$ lies on the line $y = x$, its coordinates are $(\alpha, \alpha)$, and substituting these into the distance relation $BC^2 = 4AB^2$ yields the equation $(\alpha + 2)^2 + (\alpha + 6)^2 = 4((\alpha - 4)^2 + (\alpha - 6)^2)$. Expanding and simplifying this expression results in the quadratic equation $\alpha^2 - 16\alpha + 28 = 0$, which has the solution $\alpha = 14$ when excluding the degenerate case where $B$ coincides with $D$. Calculating $\alpha + 2\beta$ with $\alpha = 14$ and $\beta = 14$ gives $14 + 2(14) = 42$.

The locus of a point whose distances from two fixed points are in a constant ratio is determined by applying the distance formula and squaring both sides of the relationship to avoid square roots. By defining the moving point as $P(x, y)$, we equate the ratio of the squared distances from $(2, 1)$ and $(1, 3)$ to the square of the given ratio, establishing the equation $\frac{(x-2)^2 + (y-1)^2}{(x-1)^2 + (y-3)^2} = \frac{25}{16}$.

Expanding this relationship involves cross-multiplying the denominators, leading to $16((x-2)^2 + (y-1)^2) = 25((x-1)^2 + (y-3)^2)$. When these expressions are fully expanded and rearranged into the standard quadratic form $ax^2 + by^2 + cxy + dx + ey + 170 = 0$, the equation simplifies to $9x^2 + 9y^2 + 14x - 118y + 170 = 0$. Comparing this with the general form identifies the coefficients as $a = 9$, $b = 9$, $c = 0$, $d = 14$, and $e = -118$.

Substituting these specific values into the required expression $a^2 + 2b + 3c + 4d + e$ leads to the calculation $9^2 + 2(9) + 3(0) + 4(14) - 118$. Performing the arithmetic results in $81 + 18 + 0 + 56 - 118$, which simplifies to $155 - 118$, yielding a final value of $37$.

Determining the equation for the side of the new triangle begins by finding the linear equation for $AC$ using points $A(-1,3)$ and $C(3,-1)$, which yields $x+y=2$. Since the new triangle shifts this side inwards by one unit, we model the parallel line as $x+y=d$ and use the distance formula $\frac{|d-2|}{\sqrt{1^2+1^2}}=1$ to solve for $d$. This leads to $|d-2|=\sqrt{2}$, giving two potential values, $d=2+\sqrt{2}$ and $d=2-\sqrt{2}$. Selecting the value that positions the line closest to the origin results in $d=2-\sqrt{2}$, which gives the final equation $x+y-(2-\sqrt{2})=0$.

This problem exploits the geometric constraints of a right isosceles triangle $OPQ$ where the vertex $O$ sits at the origin. Since $OP$ and $OQ$ are equal in length, let $OP = OQ = r$, which allows us to define the points $P$ and $Q$ in polar coordinates as $(r \cos \theta, r \sin \theta)$ and $(-r \sin \theta, r \cos \theta)$ to maintain the right angle at the origin while ensuring both points lie on the line $3x + 4y = 12$.

By substituting these coordinates into the line equation, we obtain the relations $r(3 \cos \theta + 4 \sin \theta) = 12$ and $r(-3 \sin \theta + 4 \cos \theta) = 12$. Squaring and adding these two equations simplifies the trigonometric components to $9 + 16 = 25$ based on the identity $\sin^2 \theta + \cos^2 \theta = 1$. This calculation yields $2(12/r)^2 = 25$, which determines that $r^2 = 288/25$.

Because $PQ$ is the hypotenuse of the right triangle $OPQ$, we have $PQ^2 = OP^2 + OQ^2 = 2r^2$. The expression $l = OP^2 + PQ^2 + QO^2$ therefore becomes $r^2 + 2r^2 + r^2 = 4r^2$. Substituting $r^2 = 288/25$ gives $l = 4 \times (288/25) = 1152/25 = 46.08$. Consequently, the greatest integer less than or equal to $l$ is 46.

The reflection principle effectively demonstrates that a light ray traveling from $P(1, 2)$ to $R(4, 3)$ and reflecting off the $x$-axis follows the same geometry as a straight line drawn between the image of $P$, labeled $P'(1, -2)$, and the point $R$. Calculating the slope between $P'$ and $R$ gives $\frac{3 - (-2)}{4 - 1} = \frac{5}{3}$, which establishes the line equation $y - 3 = \frac{5}{3}(x - 4)$. Setting $y = 0$ allows for the determination of the reflection point $Q$, resulting in the coordinates $\left(\frac{11}{5}, 0\right)$.

Given that $PQRS$ is a parallelogram, its diagonals $PR$ and $QS$ must bisect each other at a single midpoint. The midpoint of $PR$ is $\left(\frac{5}{2}, \frac{5}{2}\right)$. By equating this to the midpoint of $QS$, which is defined by $\left(\frac{\frac{11}{5}+h}{2}, \frac{k}{2}\right)$, the values $h = \frac{14}{5}$ and $k = 5$ are derived. The expression $(hk)^2$ is evaluated as:

$(hk)^2 = \frac{14}{5} \times 5^2 = 70$

The circumcenter of any triangle is the point where the perpendicular bisectors of the sides meet, but for a right-angled triangle, this point simplifies significantly to the midpoint of the hypotenuse. We can verify if this triangle is right-angled by comparing the slopes of the given lines $4x+3y=69$ and $4y-3x=17$. The slope of the first line is $-4/3$ and the slope of the second is $3/4$, and since their product is $-1$, these two lines are perpendicular, confirming a right angle exists at their intersection.

To determine the circumcenter $(\alpha, \beta)$, we find the endpoints of the hypotenuse by identifying the intersection points of the third line, $x+7y=61$, with the two perpendicular lines. The intersection of $x+7y=61$ and $4x+3y=69$ is the point $(12, 7)$, while the intersection of $x+7y=61$ and $4y-3x=17$ is $(5, 8)$. The circumcenter is the midpoint of the segment connecting these two vertices:
$\alpha = \frac{12+5}{2} = \frac{17}{2}, \quad \beta = \frac{7+8}{2} = \frac{15}{2}$
Calculating the final required expression with these values yields:
$(\alpha - \beta)^2 + \alpha + \beta = \left(\frac{17}{2} - \frac{15}{2}\right)^2 + \frac{17}{2} + \frac{15}{2} = 1 + 16 = 17$

Trisecting a line segment relies on the section formula to divide the span between the intercepts of the line $9x + 5y = 45$ at $(5, 0)$ and $(0, 9)$ into three equal parts. By finding the points that divide this segment in the ratios $1:2$ and $2:1$, we obtain the coordinates $(\frac{10}{3}, 3)$ and $(\frac{5}{3}, 6)$.

With these points established, calculating the slopes of the lines $l_1$ and $l_2$ that pass through the origin yields $m_1 = \frac{3 - 0}{\frac{10}{3} - 0} = \frac{9}{10}$ and $m_2 = \frac{6 - 0}{\frac{5}{3} - 0} = \frac{18}{5}$. Summing these two values gives $m_1 + m_2 = \frac{9}{10} + \frac{18}{5} = \frac{45}{10} = \frac{9}{2}$, which defines the line $y = \frac{9}{2}x$, or $9x - 2y = 0$. Determining the intersection of this line with the original $9x + 5y = 45$ involves solving the two equations together, resulting in the coordinates $(\frac{10}{7}, \frac{45}{7})$. Substituting these values into the expression $y - x$ yields $\frac{45}{7} - \frac{10}{7} = \frac{35}{7} = 5$, showing that the intersection point lies on the line defined by $y - x = 5$.

For a homogeneous second-degree equation expressed as $a x^2 + 2 h x y + b y^2 = 0$, the pair of lines representing the angle bisectors satisfies the relationship $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$. Comparing this standard form to the given equation $2 x^2 + x y - 3 y^2 = 0$, we identify the coefficients as $a = 2$, $2 h = 1$ which implies $h = \frac{1}{2}$, and $b = -3$. Substituting these specific values into the bisector formula produces $\frac{x^2 - y^2}{2 - (-3)} = \frac{xy}{1/2}$, which simplifies to $\frac{x^2 - y^2}{5} = 2 x y$. Multiplying both sides by $5$ yields the final equation $x^2 - y^2 - 10 x y = 0$.

In a parallelogram, the diagonals bisect each other at a common midpoint. Given the coordinates of vertex $A$ as $(-4, 5)$ and vertex $C$ as $(3, 2)$, the center of the parallelogram is the midpoint of diagonal $AC$, calculated as $(\frac{-4 + 3}{2}, \frac{5 + 2}{2})$, which simplifies to $(-\frac{1}{2}, \frac{7}{2})$. The other diagonal, $BD$, passes through this midpoint, and its equation is determined to be $7x + y = 0$.

The perpendicular distance $d$ from point $A(-4, 5)$ to the line representing the diagonal $BD$ is found using the formula $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$. Substituting the given values, we get $d = \frac{|7(-4) + 1(5)|}{\sqrt{7^2 + 1^2}}$, which simplifies to $\frac{|-28 + 5|}{\sqrt{49 + 1}}$, or $\frac{23}{\sqrt{50}}$. Squaring this value and multiplying by 50 yields $50d^2 = 50 \cdot (\frac{23}{\sqrt{50}})^2 = 529$.

An orthocentre is the point where the altitudes of a triangle intersect. By determining the slopes of the sides, we can identify the equations of two altitudes and solve for their intersection $(\alpha, \beta)$. For the side connecting the vertices $(1, 2)$ and $(3, 1)$, the slope is $\frac{1-2}{3-1} = -\frac{1}{2}$. Consequently, the altitude from the vertex $(2, 3)$ perpendicular to this side must have a slope of $2$. Using the point-slope form, the equation for this altitude is $y - 3 = 2(x - 2)$, which simplifies to $\beta = 2\alpha - 1$.

The side connecting the vertices $(2, 3)$ and $(3, 1)$ has a slope of $\frac{1-3}{3-2} = -2$. The altitude from the vertex $(1, 2)$ perpendicular to this side must have a slope of $\frac{1}{2}$, yielding the equation $y - 2 = \frac{1}{2}(x - 1)$, or $2\beta - 4 = \alpha - 1$. Substituting $\beta = 2\alpha - 1$ into this relationship gives $2(2\alpha - 1) - 4 = \alpha - 1$. This simplifies to $4\alpha - 2 - 4 = \alpha - 1$, which further reduces to $3\alpha = 5$, resulting in $\alpha = \frac{5}{3}$ and $\beta = \frac{7}{3}$.

With these coordinates, the roots of the required quadratic equation are $\alpha + 4\beta = \frac{5}{3} + 4\left(\frac{7}{3}\right) = \frac{33}{3} = 11$ and $4\alpha + \beta = 4\left(\frac{5}{3}\right) + \frac{7}{3} = \frac{27}{3} = 9$. Using these roots to form the equation $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$, we calculate $x^2 - (11+9)x + (11 \times 9) = 0$, which yields $x^2 - 20x + 99 = 0$.

The orthocenter of a triangle is the intersection point of its altitudes, so finding the coordinates involves determining the equations of two altitudes by finding lines that pass through a vertex and are perpendicular to the opposite side. The slope of side $BC$ connecting $(-1, 2)$ and $(4, 5)$ is $\frac{5-2}{4-(-1)} = \frac{3}{5}$. Since the altitude from vertex $A(3, -7)$ must be perpendicular to $BC$, its slope is $-\frac{5}{3}$, and the equation of this line is $y - (-7) = -\frac{5}{3}(x - 3)$, which simplifies to $5x + 3y + 6 = 0$.

Side $AC$ connects points $(3, -7)$ and $(4, 5)$, giving it a slope of 12. The altitude from vertex $B(-1, 2)$ must therefore have a slope of $-\frac{1}{12}$, resulting in the equation $y - 2 = -\frac{1}{12}(x - (-1))$, which simplifies to $x + 12y = 23$. Solving the system $5x + 3y = -6$ and $x + 12y = 23$ yields the coordinates of the orthocenter $(\alpha, \beta)$ as $\alpha = -\frac{47}{19}$ and $\beta = \frac{121}{57}$. Substituting these into the expression $9\alpha - 6\beta + 60$ gives $9(-\frac{47}{19}) - 6(\frac{121}{57}) + 60$, which simplifies to $-\frac{423}{19} - \frac{242}{19} + 60 = -\frac{665}{19} + 60 = -35 + 60$, resulting in 25.

Vertex $A$ is the intersection of $2x + y = 0$ and $x - y = 3$, giving $A(1, -2)$. With $B(\alpha, -2\alpha)$ and $C(\beta+3, \beta)$ located on their respective lines, the centroid $P(2, a)$ implies the relations $\alpha + \beta = 2$ and $\beta - 2\alpha = 3a + 2$, which simplify to $\alpha = -a$ and $\beta = a + 2$. Substituting these coordinates into the line equation $x + py = 21 a$ for both $B$ and $C$ leads to $p = 11$ and $a = 3$. Consequently, the vertices are $B(-3, 6)$ and $C(8, 5)$, and the squared length of $BC$ is $(BC)^2 = (8 - (-3))^2 + (5 - 6)^2 = 122$.

Since the points $B$ and $C$ reside on the line $y+x=0$ and are symmetric with respect to the origin, the origin serves as the midpoint of the segment $BC$. Consequently, the perpendicular bisector of $BC$ must pass through the origin and possess a slope perpendicular to the line $y=-x$, which is $m=1$. This defines the equation of the perpendicular bisector as $y=x$.

Because vertex $A$ lies on the line $y-2x=2$ and must also lie on the perpendicular bisector for the triangle to be equilateral, point $A$ is located at the intersection of $y-2x=2$ and $y=x$. Solving these equations simultaneously by substituting $x$ for $y$ yields $x-2x=2$, which simplifies to $-x=2$, giving the coordinates $(-2,-2)$ for $A$.

The altitude of the triangle, $h$, is the perpendicular distance from vertex $A$ to the line $x+y=0$. Using the standard point-to-line distance formula, $h = \frac{|-2 + (-2)|}{\sqrt{1^2+1^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$. For an equilateral triangle with height $h$, the area can be expressed as $\frac{h^2}{\sqrt{3}}$. Substituting the value $h^2 = (2\sqrt{2})^2 = 8$ into this expression results in an area of $\frac{8}{\sqrt{3}}$.